I need help in this problem ,royden 4th edition . Let $g$ be integrable over $[a,b]$ .Define the antiderivative of $g$ to be the function $f$ defined on $[a,b]$ by $f(x)=\int_{a}^{x} g~~ for~all~ x\in [a,b]$. Show that $f$ is differentiable almost everywhere on $(a,b)$.

I know from Lebesgues Theorem if the function $f$ is monotone on the open intervals $(a,b)$ , then it is differentiable almost everywhere on $(a,b)$ But integrable not always monotone .

and I can not show that if $f$ is not differentiable at $x$ then $x \in E$ and $m(E)=0$ These are the only information I have about differentiable almost everywhere

I saw your comment but I can not add comment , how can I discusses your comment with you

  • 1
    $\begingroup$ For real-valued $g$, write $f$ as the difference of two monotonic functions. $g = \lvert g\rvert - (\lvert g\rvert - g)$ $\endgroup$ – Daniel Fischer Oct 28 '15 at 19:25
  • $\begingroup$ See the fundamental theorem of calculus, and also the well-known fact that an integrable function (in the Riemann sense) is almost everywhere continuous. $\endgroup$ – Arthur Oct 28 '15 at 19:29
  • $\begingroup$ @DanielFischer Why should $g$ equal such a difference? $\endgroup$ – zhw. Oct 28 '15 at 19:41
  • $\begingroup$ Does "integrable" mean Riemann integrable? $\endgroup$ – zhw. Oct 28 '15 at 19:41
  • $\begingroup$ Whether $g$ is Lebesgue integrable or Riemann integrable we can write $g=g_1-g_2$ where $g_1$ and $g_2$ are nonnegative and integrable in the same sense. If eva had known that functions of bounded variation are a.e. differentiable then no hint would be needed. $\endgroup$ – B. S. Thomson Oct 28 '15 at 19:55

Based on the comments from @Daniel Fischer

Since $g = |{g}| - (|{g}|-g)$ and $g$ is integrable, the linearity of Lebesgue integral implies \begin{align*} f(x) = \int_a^x g = \int_a^x |{g}| - (|{g}|-g) = \int_a^x |{g}| - \int_a^x(|{g}|-g) = f_1(x)-f_2(x), \end{align*} where $f_1(x) = \int_a^x |{g}|$ and $f_2(x) = \int_a^x(|{g}|-g)$. Since $|{g}|\geq 0$ and $|{g}|-g\geq0$, the functions $f_1$ and $f_2$ are increasing on $(a,b)$. Thus, $f_1$ and $f_2$ are differentiable almost every on $(a,b)$. Since addition preserves differentiability, $f$ is differentiable a.e. on $(a,b)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.